Question

Частные производные и ещё какая-то дичь. Решите все, пожалуйста.

Answer 1

Ответ:

1.

$z = arccos(x + y)$

$\frac{dz}{dx} = - \frac{1}{ \sqrt{1 - {(x + y)}^{2} } } \times 1 = \\ = - \frac{1}{ \sqrt{1 - {x}^{2} - 2xy - {y}^{2} } }$

$\frac{dz}{dy} = - \frac{1}{ \sqrt{1 - {(x + y)}^{2} } } \times 1 = \\ = - \frac{1}{ \sqrt{1 - {x}^{2} - 2xy - {y}^{2} } }$

2.

$u = \frac{y}{x} \\ x = {e}^{t} \\ y = 1 - {e}^{2t}$

формула:

$\frac{du}{dt} = \frac{du}{dx} \times \frac{dx}{dt} + \frac{du}{dy} \times \frac{dy}{dt}$

$\frac{du}{dx} = (y {x}^{ - 1} )' = - y {x}^{ - 2} = - \frac{y}{ {x}^{2} }$

$\frac{du}{dy} = \frac{1}{x}$

$\frac{dx}{dt} = {e}^{t}$

$\frac{dy}{dt} = - 2 {e}^{2t}$

$\frac{du}{dt} = - \frac{y}{ {x}^{2} } {e}^{t} - \frac{2 {e}^{2t} }{x}$

3.

$z = \cos(3 {x}^{2} - {y}^{3} )$

$\frac{dz}{dx} = - \sin(3 {x}^{2} - {y}^{3} ) \times 6x = \\ = - 6x \sin(3 {x}^{2} - {y}^{3} )$

$\frac{dz}{dy} = - \sin(3 {x}^{2} - {y}^{3} ) \times ( - 3 {y}^{2} ) = \\ = 3 {y}^{2} \sin(3 {x}^{2} - {y}^{3} )$

4.

$z = {x}^{2} + {y}^{2} - xy + x + y$

$\frac{dz}{dx} = 2x - y + 1$

$\frac{dz}{dy} = 2y - x + 1$